1. **Problem statement:** We have a field ABCD with given sides and angles. We need to find (a) the length of CD and (b) the angle ABD. 2. **Given data:** - AB = 85 m (B north of A) - AD = 72 m - BD = 129 m - \(\angle BDC = 39^\circ\) - \(\angle BCD = 60^\circ\) 3. **Part (a): Calculate CD** - In triangle BCD, sum of angles is 180°. - Calculate \(\angle CBD = 180^\circ - 60^\circ - 39^\circ = 81^\circ\). - Use the Law of Sines: $$\frac{CD}{\sin 39^\circ} = \frac{129}{\sin 81^\circ}$$ - Solve for CD: $$CD = \frac{129 \times \sin 39^\circ}{\sin 81^\circ}$$ - Calculate values: \(\sin 39^\circ \approx 0.6293\), \(\sin 81^\circ \approx 0.9877\) - So, $$CD \approx \frac{129 \times 0.6293}{0.9877} \approx 82.2\text{ m}$$ 4. **Part (b): Show that \(\angle ABD = 31.6^\circ\)** - Consider triangle ABD. - We know AB = 85 m, AD = 72 m, BD = 129 m. - Use the Law of Cosines to find \(\angle ABD\) (angle at B): $$BD^2 = AB^2 + AD^2 - 2 \times AB \times AD \times \cos(\angle ABD)$$ - Rearranged: $$\cos(\angle ABD) = \frac{AB^2 + BD^2 - AD^2}{2 \times AB \times BD}$$ - Substitute values: $$\cos(\angle ABD) = \frac{85^2 + 129^2 - 72^2}{2 \times 85 \times 129} = \frac{7225 + 16641 - 5184}{21930} = \frac{18782}{21930} \approx 0.8565$$ - Calculate angle: $$\angle ABD = \cos^{-1}(0.8565) \approx 31.6^\circ$$ **Final answers:** - (a) \(CD \approx 82.2\text{ m}\) - (b) \(\angle ABD = 31.6^\circ\) (correct to 1 decimal place)